Increasing the concentration of the reactants increases the reaction rate (as given by Option A).
Suppose that we have to measure the rate of change of y as x changes, then we have:
[tex]Rate = \dfrac{y_2 - y_1}{x_2 - x_1}[/tex]
where we have
[tex]\rm when \: x=x_1, y = y_1\\when\: x = x_2, y= y_2[/tex]
Remember that, we divide by the change in independent variable so that we get some idea of how much the dependent quantity changes as we change the independent quantity by 1 unit.
(5 change per 3 unit can be rewritten as 5/3 change per 1 unit)
It is given that original reaction takes 10 seconds.
So rate is: One original reaction per 10 second, or one reaction/10 per 1 second.
Now, when we increase concentration, less time is taken, so we get:
New rate = One original reaction per 5 second, or one reaction/5 per 1 second.
You divide by bigger number, your output would be smaller.
[tex]5 < 10 \implies \dfrac{a}{5} > \dfrac{a}{10}[/tex] for a positive constant 'a'
Thus, the rate of reaction increases as time taken by the reaction decreases.
New rate is twice as fast as old rate, as in 10 seconds, the new rate will make two reaction to get finished, whereas old rate would've got only 1 reaction completed.
Thus, increasing the concentration of the reactants increases the reaction rate (as given by Option A).
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